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Math Help - Integration Help

  1. #1
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    Integration Help

    How would you integrate

    (2+t)^2 / 1+t^2 dt

    Had a long uni break and now i'm trying to get back into it but i've forgotten a lot.

    Please show intermediate steps. I've tried integration by parts, is that the right way to tackle it?
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  2. #2
    o_O
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    Note that:
    \begin{aligned}(2+t)^2 & = t^2 + 4t + 4 \\ & = {\color{red}t^2 + 1} + 4t + 3 \end{aligned}

    So: \frac{(2+t)^2}{1+t^2}  = \frac{{\color{red}t^2 + 1} + 4t + 3 }{1 + t^2}  = \underbrace{\frac{{\color{red} t^2 + 1}}{1+t^2}}_{=1} + \frac{4t }{1+t^2} + \frac{3}{1+t^2}

    So your integral becomes:
    \int \frac{(2+t)^2}{1+t^2} \ dt \ \ = \ \ \int \left( 1 + \frac{4t}{1+t^2} + 3 \cdot \frac{1}{1+t^2}\right) dt

    On the RHS, use a u-sub for the second integral and \arctan (x) should ring a bell for the third.
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  3. #3
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    Thanks mate

    Thanks for that, the world needs more people like yourself.

    Just did it and got

    2 ln|1+t^2| + 3tan^-1(t) + t + C
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